Parallelogram 41 • Level 4 • 12 Jun 2025Paradoxical probability

Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

a portmanteau word combining parallel and telegram. A message sent each week by the Parallel Project to bright young mathematicians.

Tackle each Parallelogram in one go. Don’t get distracted.
Finish by midnight on Sunday if your whole class is doing parallelograms.
Your score & answer sheet will appear immediately after you hit SUBMIT.
Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.

1. Intermediate Maths Challenge Problem (UKMT)

2 marks

1.1 A 24-hour digital clock shows the time in hourse and minutes.

How many times in one day will it display all four digits 2, 0, 1, and 9, in some order?

6
10
12
18
24
(Not answered)

Show Hint (–1 mark)

1 mark

The 9 can only go in the second or fourth place, so list all possibilities for each of those two cases.

The ten possible times are:

20:19
21:09
12:09
10:29
01:29
02:19
19:20
19:02
09:12
09:21

2. Intermediate Maths Challenge Problem (UKMT)

3 marks

2.1 What is the value of 2468×24682468+2468?

2
1,234
2,468
4,936
6,091,024
(Not answered)

Show Hint (–1 mark)

1 mark

Let x = 2468, then express the calculation in terms of x, and simplify.

Let x=2468, then the expression becomes x22x=x2=1234.

2 marks

2.2 What is the value of 2468×2468×24682468×2468+2468×2468?

Correct Solution: 1234

Let x=2468, then the expression becomes x32x2=x2=1234.

3. A paradoxical probability

Probability has the power to surprise, confound and utterly baffle us.

The idea of randomness, which is at the heart of probability, is one that we can easily take for granted, but needs to be handled with great care.

In this Numberphile video, Grant Sanderson (3blue1brown) explains a thought experiment by Joseph Bertrand, where choosing a chord ‘at random’ in a circle leads to three different values for the same probability!

Watch the video to see how those values come about, and what this tells us about randomness.

(If you have problems watching the video, right click to open it in a new window)

2 marks

3.1 What is the randomly selected chord being compared to in the calculation of the probability?

The radius of the circle
The circumference of the circle
The length of each side of an inscribed equilateral triangle
The length of the shorter side of an inscribed isosceles triangle
(Not answered)

2 marks

3.2 In the first example, we take any two points on the circumference of the circle.

What is the probability that the chord length is longer?

14
13
23
1
(Not answered)

Show Hint (–1 mark)

1 mark

The triangle splits the circumference into three equal segments.

Imagine a chord that starts at a vertex of the triangle - if the other end of the chord is in the opposite segment, it is longer.

Otherwise it is shorter.

2 marks

3.3 The second method gives a probability of 14 by considering a smaller circle inscribed in the triangle.

What does a longer chord length correspond to in this case?

The midpoint of the chord lying inside the smaller circle
The midpoint of the chord lying outside the smaller circle
The midpoint of the chord lying on the circumference the smaller circle
The whole chord lying outside the smaller circle
(Not answered)

2 marks

3.4 What happens when Grant runs simulations of each of the three methods?

The simulations all give the same probability
The simulations give 3 different probabilities that agree with Grant’s values
The simulations reveal an error in Grant’s methods
The simulations crash because they can’t handle the problem
(Not answered)

2 marks

3.5 What was Joseph Bertrand’s main conclusion from this thought experiment?

There’s no such thing as a probability
Maths is full of logical contradictions
Circles don’t exist
We have to be careful when dealing with probabilities involving infinitely many possible outcomes
(Not answered)

4. Intermediate Maths Challenge Problem (UKMT)

3 marks

4.1 The diagram shows six congruent equilateral triangles, of side-length 2, placed together to form a parallelogram.

What is the length of PR?

213
7
63
9
73
(Not answered)

Show Hint (–1 mark)

1 mark

Draw in a right-angled triangle and
apply Pythagoras’ Theorem.

The point Q is a vertex of the parallelogram, as shown. We extend the line PQ to the point U so that QU has length 2.

It can be seen that RQU is an equilateral triangle with side length 2.

We let T be the foot of the perpendicular from R to QU.

It can be seen that the triangles RTQ and RTU are congruent.

It follows that QT=TU=1.

By Pythagoras’ Theorem, applied to the right-angled triangle RTQ, we have RQ2=RT2+TQ2.

It follows that RT = sqrt(RQ^2 − TQ^2) = sqrt(2^2 − 1^2) = sqrt(3).

We see that PT=7.

By Pythagoras’ Theorem, applied to the right-angled triangle RTP, we have PR2=PT2+RT2=72+32=49+3=52.

Therefore PR=52=213.

5. Puzzle

1 mark

5.1 Two fathers take their sons fishing in the same boat.

What is the least number of possible people in the boat?

Correct Solution: 3

If they were all different people, there would be 4 in total.

But if one of them is the grandfather, then the father counts as both a father and a son, so there are only 3 in total: a boy, his father, and his grandfather.

Before you hit the SUBMIT button, here are some quick reminders:

You will receive your score immediately, and collect your reward points.
You might earn a new badge... if not, then maybe next week.
Make sure you go through the solution sheet – it is massively important.
A score of less than 50% is ok – it means you can learn lots from your mistakes.
The next Parallelogram is next week, at 3pm on Thursday.
Finally, if you missed any earlier Parallelograms, make sure you go back and complete them. You can still earn reward points and badges by completing missed Parallelograms.

Cheerio, Simon.